The lifespans of tigers in a particular zoo are normally distributed. The average tiger lives $24$ years; the standard deviation is $3.6$ years. Use the empirical rule (68-95-99.7%) to estimate the probability of a tiger living longer than $20.4$ years.
Answer: $24$ $20.4$ $27.6$ $16.8$ $31.2$ $13.2$ $34.8$ $68\%$ $16\%$ $16\%$ We know the lifespans are normally distributed with an average lifespan of $24$ years. We know the standard deviation is $3.6$ years, so one standard deviation below the mean is $20.4$ years and one standard deviation above the mean is $27.6$ years. Two standard deviations below the mean is $16.8$ years and two standard deviations above the mean is $31.2$ years. Three standard deviations below the mean is $13.2$ years and three standard deviations above the mean is $34.8$ years. We are interested in the probability of a tiger living longer than $20.4$ years. The empirical rule (or the 68-95-99.7 rule) tells us that $68\%$ of the tigers will have lifespans within 1 standard deviation of the average lifespan. The remaining $32\%$ of the tigers will have lifespans that fall outside the shaded area. Because the normal distribution is symmetrical, half $({16\%})$ will live less than $20.4$ years and the other half $({16\%})$ will live longer than $27.6$ years. The probability of a particular tiger living longer than $20.4$ years is ${68\%} + {16\%}$, or $84\%$.